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Based on my previous question here, as suggested I'm using a push-pull driver circuit to control a MOSFET gate. this is the circuit:

SCH

And this is a simulation result:

oscilloscope_simulated

  • Vertical Divisions are 1 V, Horizontal Divisions are ~7 uS.
  • The yellow pulse is from the MCU (3.3 V, T.Rise = T.Fall = 100 nS).
  • The blue curve is the optocoupler output. (Peaks at ~11.7 V).
  • The red one is the voltage at output of push-pull transistors. (Vp: ~11.2 V).

The turn on is all ok (hopefully not just in simulation), but it takes around 60 uS for the gate voltage to fall to ~1 V. The turn off timing is important for the procedure I'm trying to follow. (Ideally to be less than 10 uS.)

Question: How can the turn off times be shortened? (Which components are more effective here to adjust so the tail at turn off is less extended?)

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2 Answers 2

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Let's summarise the switching speed as it now stands:
Turn on: delay is short, and Vgs rise time is fast, so all good.
Turn-off: delay is long, and Vgs fall time is slow, which needs improving.

Delay: defined as the time from input changing by 50% to output changing by 10%. Based on the waveforms provided, it seems the delay time is about 0.5 divs, so:
Td(off) =0.5div x 7us/div = 3.5us.

This is much smaller than the total turn-off delay time, so I would ignore this for now.

Fall time: defined as the time taken for output to change from 10 % to 90%. Based on the waveforms provided, it seems this is about 4.5 divisions, so:
4.5div x 7us/div = 31.5us.

To speed this up, try the following changes, then advise the new waveforms:
Change 1: R6: reduce from 200Ω to 22Ω.
Change 2: R4: reduce from 3k3 to 100Ω.
Change 3: R5: change from 3k3 to 1k.
Change 4: R7: New component, connected from Q2 base to Q2 emitter. Value 470Ω.
Change 5: C1: New component, connected across R1. Value 200pF.

There are other improvements that are possible, but these may not be required, it depends on what results you get after making the suggested changes.

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  • \$\begingroup\$ I tried these changes in the suggested order. 1 and 2 didn't change the tail durance much. but reducing R5 helped a lot, now the gate voltage falls to 1V in 30uS after the MCU rise. that's 20uS after the MCU signal has fallen to Zero. about change4, you sure the resistor should be connected between base and emitter of the same BJT? doesn't it make the BJT switching operation bypassed ? \$\endgroup\$ Commented Dec 2, 2023 at 11:10
  • \$\begingroup\$ @Bikay Change 4: this allows Vgs to discharge below the Vbe of Q3, but need to select R7 value low enough to get the benefit, but not too low that it "shorts out" the B-E of Q2 & Q3. What is the MOSFET in your simulation? What is the waveform of opto-coupler collector current? \$\endgroup\$ Commented Dec 2, 2023 at 11:25
  • \$\begingroup\$ Shorting B-E of Q2 and Q3 or just Q2? (Cuz you said to put the resistor on Q2, is it a typo or you meant that actually?). Mosfet will be IRBP072N15 (Ciss~5400pF) but the simulation is done using IRF064 (with Ciss~7000pF) i know the simulation will show slower timings, due to higher gate capacitance ( is this assumption right?) But thought the real world is gonna be a little slower than the simulations also. \$\endgroup\$ Commented Dec 2, 2023 at 11:32
  • \$\begingroup\$ R7 will connect across B-E of both Q2 and Q3. That MOSFET (IRBP072N15) is quite large (150V 7.2mΩ), do you need such a large MOSFET? What is the load? \$\endgroup\$ Commented Dec 2, 2023 at 11:37
  • \$\begingroup\$ There will be 10 of this circuit, each an output from MCU pin, controlling mosfets with very low duty cycle pulses (frequency is 50 to 1000Hz). Some of the loads are 24V, some are 36, and a few are 80V. All pure inductive loads. (The protections at load side are considered, that's why I didn't draw them at the schematics) that's why I've chosen such big mosfets to withstand possible back EMI from large coils. (And the grounds are also separated, and the 12V rails is from a buck converter from the same HV source that supplies the loads) \$\endgroup\$ Commented Dec 2, 2023 at 12:23
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Optocouplers with phototransistors have a (too) slow switch off time.

I would put a Schmitt-Trigger between the phototransistor and the MOSFET's Gate.

There are integrated circuits like the 40106 - or discrete possibilities with bipolar transistors etc. as well.

The Schmitt trigger decreases the drop time (and rise time) drastically.

Despite that, there will be a delay from the drop of the input signal until the MOSFET's Gate voltage changes, this cannot be eliminated. This is due to the slow switching off of any phototransistor.

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  • \$\begingroup\$ It's spelled phototransistor in English, though I think most people will understand what you mean. \$\endgroup\$ Commented Dec 2, 2023 at 0:05
  • \$\begingroup\$ @Hearth Thanks, I edited it. \$\endgroup\$ Commented Dec 7, 2023 at 22:03

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